Derivative help

**Glitch** · June 11th 2010, 05:43 AM

Question:

f(x) = ln(cosx)

I need to find the derivative.

I'm unsure where to start. I tried the chain rule, and I get stuck, as follows:

$\frac {1}{x} (cosx)^0 (-sinx)$

$\frac {-sinx}{x}$

The answer is supposed to be -tanx

**pickslides** · June 11th 2010, 05:48 AM

Originally Posted by Glitch

f(x) = ln(cosx)

I need to find the derivative.

make $u = \cos x \implies y = \ln u$

$\frac{du}{dx} = -\sin x$ and $\frac{dy}{du}= \frac{1}{u}$

$<br /> \frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx}= \frac{1}{u}\times -\sin x = \frac{-\sin x}{\cos x} = -\tan x <br />$

**Glitch** · June 11th 2010, 05:52 AM

Interesting. So that's the chain rule? Your working out isn't very familiar to me.

**Raoh** · June 11th 2010, 06:04 AM

Originally Posted by Glitch

Interesting. So that's the chain rule? Your working out isn't very familiar to me.

hi

if u have two functions of $x$ , $f$ and $g$ ,how do you find the derivative of $f(g(x))$ ?

**Glitch** · June 11th 2010, 06:11 AM

I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.

**Raoh** · June 11th 2010, 06:22 AM

Originally Posted by Glitch

I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.

well ..
$\left (\ln f(x) \right )'=\frac{f'(x)}{f(x)}$
$\left (\ln \cos(x) \right )'=\frac{-\sin(x)}{\cos(x)}$
$(\cos(x))'=-\sin(x)$

**Glitch** · June 11th 2010, 06:26 AM

Yeah, I made an error. Thanks for the help!

**chisigma** · June 18th 2010, 06:48 AM

A general formula for the 'logarithmic derivative' of a function is...

d/dx ln f(x) = f'(x)/f(x) (1)

... and here is f(x) = cos x , so that...

Kind regards

$\chi$ $\sigma$

**TheCoffeeMachine** · June 18th 2010, 07:58 AM

Originally Posted by Glitch

I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.

Why were you lowering the power, though? The derivative of ln{x} is 1/{x} and the derivative of cosx is -sinx. That should be sufficient:

$\dfrac{d}{dx}\left\{\ln(\cos{x})\right\} = \left(\dfrac{1}{\cos{x}}\right)(-\sin{x}) = -\dfrac{\sin{x}}{\cos{x}} = -\tan{x}.$

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